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Scalar field

  • #1

Homework Statement



This is not really a problem but I was going over my lecture notes and I see [itex] \mathscr{H}=\frac{1}{2}\left(\pi^{2} + \vec{\nabla}\phi \cdot \vec{\nabla}\phi + m^{2}\phi^{2}\right) [/itex] and [itex]\frac{\partial\mathscr{H}}{\partial\phi} = -\nabla^{2}\phi + m^{2}\phi [/itex]

Homework Equations





The Attempt at a Solution



I would think that [itex]\frac{\partial\mathscr{H}}{\partial\phi} = \nabla^{2}\phi + m^{2}\phi [/itex]. But I don't know where the minus sign is coming from.
 

Answers and Replies

  • #2
I just found a vector identity [itex] \vec{\nabla}\phi\cdot\vec{\nabla}\phi = \vec{\nabla}\cdot\left(\phi\vec{\nabla}\phi\right) - \phi\vec{\nabla}^{2}\phi [/itex]. I now see how the result follow.

EDIT: I'm confused again. will the phi derivative of the first term vanish?
 
  • #3
If by "the first term" you mean the [itex]\pi^2[/itex] term, then yes, the [itex]\phi[/itex] derivative will kill that term; reason being [itex]\phi[/itex] doesn't appear in that term, only its time derivative.

Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.
 
Last edited:
  • #4
dextercioby
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The 3-divergence in the Hamiltonian density (i.e. the first term in the RHS of the equality in post #2) can be discarded, since by integration of full space gives 0, because scalar fields are normally taken from the Schwartz space of test functions.
 
  • #5
The 3-divergence in the Hamiltonian density (i.e. the first term in the RHS of the equality in post #2) can be discarded, since by integration of full space gives 0, because scalar fields are normally taken from the Schwartz space of test functions.
If by "the first term" you mean the [itex]\pi^2[/itex] term, then yes, the [itex]\phi[/itex] derivative will kill that term; reason being [itex]\phi[/itex] doesn't appear in that term, only its time derivative.

Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.


That makes sense. I also realized that it can be integrated by parts using the same reasoning of vanishing at the boundaries. Thanks!
 

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